Question

An ellipse has its centre at (1, −1) and semi-major axis = 8 and it passes through the point (1, 3). The equation of the ellipse is

पर्याय

• \[\frac{\left( x + 1 \right)^2}{64} + \frac{\left( y + 1 \right)^2}{16} = 1\]

   

• \[\frac{\left( x - 1 \right)^2}{64} + \frac{\left( y + 1 \right)^2}{16} = 1\]

   

• \[\frac{\left( x - 1 \right)^2}{16} + \frac{\left( y + 1 \right)^2}{64} = 1\]

   

• \[\frac{\left( x + 1 \right)^2}{64} + \frac{\left( y - 1 \right)^2}{16} = 1\]

   

Answer 1

\[\frac{\left( x - 1 \right)^2}{64} + \frac{\left( y + 1 \right)^2}{16} = 1\]
\[\text{ According to the question, the centre is at }\left( 1, - 1 \right).\]
\[a = 8 \left(\text{ Given }\right)\]
\[ \Rightarrow \frac{\left( x - 1 \right)^2}{a^2} + \frac{\left( y + 1 \right)^2}{b^2} = 1 . . . (1)\]
\[\text{ It passes through point }\left( 1, 3 \right).\]
\[i . e . x = 1\text{ and }y = 3\]
Putting these values in eq . (1), we get:
\[\frac{\left( 1 - 1 \right)^2}{a^2} + \frac{\left( 3 + 1 \right)^2}{b^2} = 1\]
\[ \Rightarrow \frac{16}{b^2} = 1\]
\[ \Rightarrow b^2 = 16 or b = 4\]
Substituting the values ofaandbin eq. (1), we get:
\[\frac{\left( x - 1 \right)^2}{64} + \frac{\left( y + 1 \right)^2}{16} = 1\]